5 条题解
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219220448 董文杰 (董文杰) LV 9 @ 2022-10-05 23:11:19
//罕见的C程序题解 #include<stdio.h> int main() { int n; int a[100]; scanf("%d", &n); for (int i = 0; i < n; i++) scanf("%d", &a[i]); double sum1 = 0,sum2 = 0,sum3=0,sum4=0; for (int i = 0; i < n; i++) { if (a[i] >= 0 && a[i] <= 18) sum1++; else if (a[i] >= 19 && a[i] <= 35) sum2++; else if (a[i] >= 36 && a[i] <= 60) sum3++; else if (a[i] >= 61 ) sum4++; } printf("%.2f%%\n%.2f%%\n%.2f%%\n%.2f%%\n", 100*sum1 / n, 100*sum2 / n, 100*sum3 / n, 100*sum4 / n); return 0; }
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12024-08-28 11:55:55@
#include<stdio.h> using namespace std; int cnt[4] = {0, 0, 0, 0}, n, q; int main(){ scanf("%d", &n); for(int i = 1; i <= n; i++){ scanf("%d", &q); if(q <= 18)cnt[0]++; else if(q >= 19 && q <= 35)cnt[1]++; else if(q >= 36 && q <= 60)cnt[2]++; else cnt[3]++; } for(int i = 0; i < 4; i++)printf("%.2f%%\n", float(cnt[i])/n*100); return 0; }
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12022-07-28 10:24:49@
嗨嗨嗨
#include <bits/stdc++.h>
using namespace std;
int main()
{
int n,m,a,s1=0,s2=0,s3=0,s4=0;
cin>>n;
m=n;
while(n--)
{
cin>>a;
if(a>=61)
s1++;
else if(a>=36)
s2++;
else if(a>=19)
s3++;
else
s4++;
}
printf("%.2lf%\n%.2lf%\n%.2lf%\n%.2lf%\n",100.0*s4/m,100.0*s3/m,100.0*s2/m,100.0*s1/m);return 0;
} -
02024-11-15 18:46:13@
#include<stdio.h>
int main()
{
int n;
int a[100];
scanf("%d", &n);
for (int i = 0; i < n; i++)
scanf("%d", &a[i]);
double sum1 = 0,sum2 = 0,sum3=0,sum4=0;
for (int i = 0; i < n; i++)
{
if (a[i] >= 0 && a[i] <= 18)
sum1++;
else if (a[i] >= 19 && a[i] <= 35)
sum2++;
else if (a[i] >= 36 && a[i] <= 60)
sum3++;
else if (a[i] >= 61 )
sum4++;
}
printf("%.2f%%\n%.2f%%\n%.2f%%\n%.2f%%\n", 100*sum1 / n, 100*sum2 / n, 100*sum3 / n, 100*sum4 / n);return 0;
} -
02022-07-31 15:41:48@
很好,很有精神!
- 1
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